Is (x^2 + y^2) a perfect square? (1) x is even, y is odd (2)

Nice work Version2. Yes, sum of perfect squares being themselves a perfect square must have a root which is divisible by 4:), Hence an even root

ie (3,4) (5,12) (12,16)

halle. do you have the answer to this?

Can I ask someone to provide a solution for this? I don't understand the solution provided by Paul. Thank you.

Is (x^2 + y^2) a perfect square?

(1) x is even, y is odd --> clearly insufficient: if \(x=0\) and \(y=1\) then \(x^2+y^2=1^2\) and the answer will YES but if \(x=2\) and \(y=1\) then \(x^2+y^2=5\) and the answer will be NO. Not sufficient.

(2) x and y are odd --> \(x=2m+1\) and \(y=2n+1\) for some integers \(m\) and \(n\) --> \(x^2+y^2=(2m+1)^2+(2n+1)^2=2*(2(m^2+m+n^2+n)+1)=2*odd\) (as \(2(m^2+m+n^2+n)+1=even+odd=odd\)), so \(x^2+y^2\) is not a perfect square (an even perfect square must be multiple of 2^2=4, in other words \(x^2+y^2\) to be a perfect square since it's a multiple of 2 it must be multiple of 2^2 as a perfect square has even powers of its primes, but as \(x^2+y^2=2*odd\) then it's not a multiple of 2^2). Sufficient.

But even though formal answer to the question is B (Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient), this is not a realistic GMAT question, as: on the GMAT, two data sufficiency statements always provide TRUE information and these statements never contradict each other, and in this question (1) says that x is even and (2) says that x is odd, so the statements clearly contradict each other.

Hope it's clear.